JEE Advanced 2020 · Paper 1 · Physics Q6

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PhysicsSingle correct+3 / -1Question 6 of 54

An open-ended U-tube of uniform cross-sectional area contains water (density $10^3\ \mathrm{kg\,m^{-3}}$ ). Initially the water level stands at 0.29 m from the bottom in each arm. Kerosene oil (a water-immiscible liquid) of density $800\ \mathrm{kg\,m^{-3}}$ is added to the left arm until its length is 0.1 m, as shown in the schematic figure below. The ratio $\left(\dfrac{h_1}{h_2}\right)$ of the heights of the liquid in the two arms is

A U-tube open at both ends. In the left arm a 0.1 m column of kerosene (black) sits on top of the water; the total liquid height from the bottom in the left arm is h₁. In the right arm the water stands at height h₂, lower than h₁.

\dfrac{15}{14}

\dfrac{35}{33}

\dfrac76

\dfrac54

Answer and worked solution

Official answer: B

Let the water surface in the left arm fall by $x$ ; the right arm then rises by $x$ .

Pressure balance at the level of the kerosene-water interface:
$\rho_kgh_k=\rho_wg(2x)\implies800\times0.1=1000\times2x\implies x=0.04\text{ m}.$

Heights.

$h_1=0.29-0.04+0.10=0.35$ m.
$h_2=0.29+0.04=0.33$ m.

$\frac{h_1}{h_2}=\frac{35}{33}.$
(B)

Topics:Pressure and Pascal's law

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JEE Advanced 2020 · Paper 1, Physics question 6